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Let $a, b \in \mathbb{R}$. Show that $|ab| \le \frac 12 (a^2 + b^2 )$.

Need consider three cases:
(i) both $a,b \ge 0$: $|ab| = ab=\frac{2ab}2= \frac{(a+b)^2-a^2-b^2}2=\frac{(a+b)^2-(a^2+b^2)}2$.

(ii) both $a,b \lt 0$: $|ab| = ab$ , same as in (i).

(iii) either one of $a,b \lt 0$: $|ab| = -ab=-\frac{2ab}2= -(\frac{(a+b)^2-a^2-b^2}2)=-(\frac{(a+b)^2-(a^2+b^2)}2)$.


So, need prove $2ab \le a^2+b^2$ which is again back to the question.

I have in mind only an approach based on triangle formed by three vectors, but am unclear about using that. The reason is need to constrain the angle formed to $90^o$, as only then $a^2 + b^2 = a^2+b^2$.
But, even with that constraint, the issue is that the vector approach has equivalent formulation only in terms of complex quantities.

So, no progress possible.

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    $\begingroup$ If the reason for closing is it being 'duplicate', then it should apply 'also' to the post causing my post's closure; as there also it is declared as duplicate. Else, the reason should be made clear. $\endgroup$ – jiten Jun 1 at 8:26
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    $\begingroup$ Indeed, it should. A big problem I see here that even though you posted this as a proof-verification question, there is A) no proof to be verified, and B) the answerers treated it as a duplicate, and reproduced the same proof as in those old threads. To a great extent your answerers prove that this is a duplicate in spite of your attempt. $\endgroup$ – Jyrki Lahtonen Jun 1 at 8:27
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    $\begingroup$ Anyway, we have more than 10 threads dedicated to essentially the same inequality. If a new users recognized it as something they can prove, and jumps to jot a proof down, I understand why that may happen. But a user with more than 20k rep should know better. $\endgroup$ – Jyrki Lahtonen Jun 1 at 8:32
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    $\begingroup$ It would have been fine, if only the answerers studied your suggested line of looking at triangles. But, no, they wanted to grab a low-hanging fruit. Actually the law of cosines would allow you to complete that approach. If the claimed inequality were not true, then selecting a suitable angle between the sides of lengths $a$ and $b$ would lead to the absurd situation that the squared length of $c$ would be negative. Let's wait for the other voters to tell their opinion. I would be fine with someone posting an answer along those lines, but getting the current answers deleted. $\endgroup$ – Jyrki Lahtonen Jun 1 at 8:46
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    $\begingroup$ If $2|ab|>a^2+b^2$ then there exists an angle $\theta, 0<\theta<\pi$ such that $$a^2+b^2-2ab\cos\theta<0.$$ Gotta rush, sorry. $\endgroup$ – Jyrki Lahtonen Jun 1 at 11:17
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We focus on $a, b > 0$.

Suppose on the contrary that we have

$$2ab > a^2 + b^2$$

but from the cosine rule, we have $$a^2 + b^2 - 2ab\cos \theta \ge 0$$

In particular, let $\cos \theta = 1$, then

$$a^2+b^2 \ge 2ab$$

which is a contradiction.

Old answer:

We have $$(|a|-|b|)^2 \ge 0$$

$$|a|^2 - 2|a||b| + |b|^2 \ge 0$$

$$a^2 - 2|a||b| + b^2 \ge 0$$

$$2|a||b| \le a^2 + b^2$$

Dividing by $2$,

$$|a||b| \le \frac12 (a^2 + b^2)$$

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  • $\begingroup$ Have another smallish question like this, from the same source: Let $a, b \in \mathbb{R}$. Show that $| |a| − |b| | \le |a − b|$. I hope here need consider all four cases: (i) $a\ge 0, b\ge 0$, (ii) $a\ge 0, b\lt 0$, (iii) $a\lt 0, b\ge 0$, (i) $a\lt 0, b\lt 0$. For (i) $|a-b|=|a-b|$, (ii) $|a+b|\lt |a-b|$, & so on. $\endgroup$ – jiten Jun 1 at 7:43
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    $\begingroup$ $|a| =|a-b+b|\le |a-b|+ |b|$, so we have $|a|- |b| \le |a-b|$ also, you can swith the role of $a$ and $b$ and conclude that. Nothing wrong with consider cases but it's good to have a shorter proof sometimes. This is called the reverse triangle inequality. $\endgroup$ – Siong Thye Goh Jun 1 at 7:48
  • $\begingroup$ btw, let me make life easier. just unaccept this solution and I will delete the answer. $\endgroup$ – Siong Thye Goh Jun 1 at 9:16
  • $\begingroup$ I amjust leaving the answer here for $15$ minutes. To be deleted. $\endgroup$ – Siong Thye Goh Jun 1 at 9:58
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    $\begingroup$ regardless of whether you viewed as vector or complex number, I just used a result that uses the length. $\endgroup$ – Siong Thye Goh Jun 1 at 10:12
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As you've already noted, we can restate the claim as $a,\,b\ge 0\implies 2ab\le a^2+b^2$. This follows from $a^2+b^2-2ab=(a-b)^2\ge0$.

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